Question

Determine the eigenvalues and eigenfunctions of the following regular Sturm- Liouville systems: (a) \(y^{\prime \prime}+\lambda y=0\), \(y(0)=0, \quad y(\pi)=0\) (b) \(y^{\prime \prime}+\lambda y=0\) \(y(0)=0, \quad y^{\prime}(1)=0\) (c) \(y^{\prime \prime}+\lambda y=0\), \(y^{\prime}(0)=0, \quad y^{\prime}(w)=0\) (d) \(y^{\prime \prime}+\lambda y=0\), \(y(1)=0, \quad y(0)+y^{\prime}(0)=0 .\)

Short answer

In summary, we have found the eigenfunctions and eigenvalues for the four different Sturm-Liouville systems as follows: 1. Case (a): Eigenfunctions: \(y_n(x) = \sin(nx)\) with \(n = 1,2,3,\dots\); Eigenvalues: \(\lambda_n = -n^2\). 2. Case (b): Eigenfunctions: \(y_n(x) = \sin(\frac{(2n-1)\pi x}{2})\) with \(n = 1,2,3,\dots\); Eigenvalues: \(\lambda_n = \frac{(2n-1)^2\pi^2}{4}\). 3. Case (c): Eigenfunctions: \(y_n(x) = \sin((2n-1)\frac{\pi x}{2w})\) with \(n = 1,2,3,\dots\); Eigenvalues: \(\lambda_n = \frac{(2n-1)^2\pi^2}{4w^2}\). 4. Case (d): Eigenfunctions: \(y_n(x) = \sin(\frac{n\pi x}{2})\) with \(n = 1,2,3,\dots\); Eigenvalues: \(\lambda_n = \frac{n^2\pi^2}{4}\).

Step-by-step solution

Apply Boundary Conditions

First, let's apply the given boundary conditions to the general equation: \(y(0)=0\) and \(y(\pi)=0\).

Solve the Differential Equation

Depending on the sign of \(λ\), the general solution of the given differential equation can have three different cases: \(\lambda < 0\), \(\lambda = 0\), and \(\lambda > 0\). (a) For \(\lambda < 0\), let \(λ = -k^2\) with \(k > 0\). Then the general solution is \(y(x) = A \cos(kx) + B \sin(kx)\). Applying the boundary conditions, \(y(0) = A \cos(0) + B \sin(0) = 0 \Rightarrow A = 0\). Next, apply the second boundary condition: \(y(\pi) = B \sin(k\pi) = 0\). Since \(B=0\) is a trivial solution, we need non-trivial values. This means that \(k\pi\) must be a multiple of \(\pi\), and \(k=n\) where \(n=1,2,3,\dots\). Therefore, the eigenfunctions are \(y_n(x) = \sin(nx)\) with \(n = 1,2,3,\dots\). Eigenvalues are \(\lambda_n = -n^2\). (b) For \(\lambda = 0\), the general solution is \(y(x) = A + Bx\). We can find that there is only a trivial solution for these boundary conditions which is not relevant for eigenfunctions. (c) For \(\lambda > 0\), let \(λ = k^2\) with \(k > 0\). Then the general solution is \(y(x) = A \cosh(kx) + B \sinh(kx)\). However, this case doesn't satisfy the given boundary conditions with non-trivial solutions. Thus, we find that Case (a) is the only relevant solution for this problem. The eigenfunctions are \(y_n(x) = \sin(nx)\) with \(n = 1,2,3,\dots\), and the eigenvalues are \(\lambda_n = -n^2\). #Case (b):# Follow the same step as Case (a): (a) For negative \(\lambda\), we get only trivial solutions. (b) For \(\lambda = 0\), the general solution is \(y(x) = A + Bx\). Applying the first boundary condition: \(y(0) = A + B(0) = 0 \Rightarrow A = 0\). So the solution becomes \(y(x) = Bx\). Then apply \(y'(1) = 0\): \(y'(x) = B \Rightarrow B = 0\). This results in a trivial solution. (c) For positive \(\lambda\), the general solution is \(y(x) = A \cos(kx) + B \sin(kx)\). Applying the boundary conditions and solving the problem: \(y_n(x) = \sin(\frac{(2n-1)\pi x}{2})\) with \(n = 1,2,3,\dots\), and the eigenvalues are \(\lambda_n = \frac{(2n-1)^2\pi^2}{4}\). #Case (c):# Using Step 1 and Step 2: (a) For \(\lambda < 0\), only trivial solutions. (b) For \(\lambda = 0\), only trivial solutions. (c) For \(\lambda > 0\), the general solution is \(y(x) = A \cos(kx) + B \sin(kx)\). Applying the boundary conditions: \(y_n(x) = \sin((2n-1)\frac{\pi x}{2w})\) with \(n = 1,2,3,\dots\), and the eigenvalues are \(\lambda_n = \frac{(2n-1)^2\pi^2}{4w^2}\). #Case (d):# Using Step 1 and Step 2: (a) For \(\lambda < 0\), only trivial solutions. (b) For \(\lambda = 0\), only trivial solutions. (c) For \(\lambda > 0\), the general solution is \(y(x) = A \cos(kx) + B \sin(kx)\). Applying the boundary conditions: \(y_n(x) = \sin(\frac{n\pi x}{2})\) with \(n = 1,2,3,\dots\), and the eigenvalues are \(\lambda_n = \frac{n^2\pi^2}{4}\).

Key concepts

Eigenvalues

In Sturm-Liouville theory, eigenvalues are special scalars associated with a differential equation that determine specific solutions. These values are crucial because they are linked to physical properties in many engineering and physics problems.
In our exercise, the eigenvalue comes from solving the differential equation: \[ y'' + \lambda y = 0 \] under different boundary conditions. Using boundary conditions, like \(y(0) = 0\) and \(y(\pi) = 0\), helps us identify meaningful solutions.
To find eigenvalues, we examine different cases for \(\lambda\):

• **Case 1:** If \(\lambda < 0\), we set \(\lambda = -k^2\). This transforms our function into trigonometric form, revealing solutions tied to negative values \(\lambda_n = -n^2\).
• **Case 2:** If \(\lambda = 0\), the solutions are often trivial, meaning they provide no useful eigenvalue.
• **Case 3:** If \(\lambda > 0\), the solutions become exponential functions, which frequently do not yield non-trivial eigenvalues under standard boundary settings.

Eigenvalues essentially give us a quantitative measure of frequency or decay in applied problems, such as vibrations and oscillations.

Eigenfunctions

Eigenfunctions are the solutions corresponding to eigenvalues for a given differential equation in the Sturm-Liouville problem.
They represent the shapes or patterns of the system's response and are essential for understanding the behavior of physical systems.
When you solve \(y'' + \lambda y = 0\) on a specific interval, you need the proper conditions:

• **Trigonometric form:** Solutions like \(y_n(x) = \sin(nx)\) appear when \(\lambda < 0\).
• **Linear form:** When \(\lambda = 0\), the general solution \(y(x) = A + Bx\) usually results in trivial eigenfunctions due to the boundary conditions.
• **Exponential form:** Settings where \(\lambda > 0\) involve \(y(x) = A \cosh(kx) + B \sinh(kx)\), however, non-trivial results are uncommon.

Eigenfunctions form the basis upon which more complex solutions or functions can be expanded, especially in contexts such as signal processing and quantum mechanics.

Boundary Conditions

Boundary conditions are crucial to solving differential equations, as they define the specific behavior of a function at the endpoints of an interval.
They help us find suitable solutions, eigenvalues, and eigenfunctions based on the specific requirements of the system.
For instance, in an equation like \(y'' + \lambda y = 0\), boundary conditions such as

• \(y(0) = 0\) and \(y(\pi) = 0\)
• \(y(0) = 0\) and \(y'(1) = 0\)
• and others

These are set to identify non-trivial solutions, ensuring the resulting function aligns with physical or theoretical expectations.
Properly applying these conditions is essential in yielding meaningful eigenvalues and eigenfunctions, describing all possible states or modes of the system.